find the integral: 1. [x - (5/x) -e ^-2x]dx 2. (-sinx / cosx)dx 3. (2t - 9)^4 dt 4. (6x - 10) / (x^2 - 2x -3) dxfor equation 4...

Because the 4th integral is more elaborated, I''ll start
by solving it:

We notice that if we note the
denominator

x^2 - 2x -3 = t => (2x-2)dx =
dt

The numerator of the ratio is 6x -
10.

The numerator could be written
as:

6x - 10 = 2x-2 + 4x-8

The
ratio will be written:

(6x - 10) / (x^2 - 2x
-3)=(2x-2)/(x^2 - 2x -3)+(4x-8)/(x^2 - 2x -3)

Int(6x - 10)
dx/ (x^2 - 2x -3)=Int(2x-2)dx/(x^2 - 2x -3)+Int(4x-8)dx/(x^2 - 2x
-3)

We'll solve Int(2x-2)dx/(x^2 - 2x
-3)

Int(2x-2)dx/(x^2 - 2x -3) = Int dt/t = ln t +
C

Int(2x-2)dx/(x^2 - 2x -3) = ln (x^2 - 2x
-3) + C

We'll solve Int(4x-8)dx/(x^2 - 2x
-3).

Int(4x-8)dx/(x^2 - 2x -3) = ln (x^2 - 2x -3) -4 Int
dx/(x^2 - 2x -3)

We'll solve Intdx/(x^2 - 2x
-3)

We'll write the denominator as a product of linear
factors.

The roots of the equation x^2 - 2x -3 = 0
are:

x1+x2 = -2

x1*x2 =
-3

x1 = -3 and x2 = 1

The
equation will be written:

 x^2 - 2x -3 = (x-x1)(x-x2) =
(x+3)(x-1)

Now, we'll write the
ratio:

1/ (x^2 - 2x -3) =
1/(x+3)(x-1)

as a sum of simple
irreducible ratios.

1/(x+3)(x-1) = a/(x+3) +
b/(x-1)

1 = a(x-1) +
b(x+3)

We'll remove the
brackets:

1 = ax - a + bx +
3b

We'll combine like terms:

1
= x(a+b) + 3b-a

a+b = 0 => -a =
b

3b-a = 1 => 3b+b = 1 => 4b = 1 =>
b = 1/4 => a =
-1/4

1/(x+3)(x-1) = -1/4(x+3) +
1/4(x-1)

Int dx/(x+3)(x-1) = (-1/4)*Int dx/(x+3) +
(1/4)*Intdx/(x-1)

Int dx/(x+3) = ln (x+3) +
C

Intdx/(x-1) = ln (x-1) +
C

Int dx/(x+3)(x-1) = (1/4)[ln (x-1) -  ln (x+3)] +
C

Int dx/(x+3)(x-1) = (1/4)*ln [(x-1)/(x+3)]  +
C

Int dx/(x+3)(x-1) = ln [(x-1)/(x+3)] +
C

The result of integral
is:

Int (6x - 10) dx/ (x^2 - 2x -3) = 3ln (x^2 - 2x -3) -
ln [(x-1)/(x+3)] + C

Int (6x - 10) dx/ (x^2 -
2x -3) = ln [(x+3)*(x^2 - 2x -3)^3/(x-1)] + C